Question

A 4.75 kg block is sent up a ramp inclined at an angle ????=31.5° from the horizontal. It is given an initial velocity ????0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is ????k=0.368 and the coefficient of static friction is ????s=0.663. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

Answer 1

Answer:

d = 13.7 m

Explanation:

When block is moving upwards along the inclined plane

then the block is decelerated due to gravity as well as due to friction and speed of the block by which it is projected upwards is given

$v_i = 15 m/s$

deceleration caused to the block due to net force opposite to the motion is given as

$F = - mg sin\theta - \mu mgcos\theta$

$a = \frac{F}{m}$

$a = -g(sin\theta + \mu cos\theta)$

since block is sliding on the inclined plane

so here we can say that the coefficient of the friction must be kinetic friction here

$a = -9.81(sin31.5 + 0.368 cos31.5)$

$a = - 8.2 m/s^2$

now for finding the distance upto which it will stop is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 15^2 = 2(-8.2) d$

$d = 13.7 m$

Answer 2

The distance traveled by the block before stopping is 13.72 m.

The given parameters;

• mass of the block, m = 4.75 kg

• angle of inclination of the ramp, Ф = 31.5⁰

• initial velocity of the block, u = 15 m/s

• coefficient of kinetic friction, $\mu_k$ = 0.368

• coefficient of static friction, $\mu_s$ = 0.663

The deceleration of the block is calculated by apply Newton's second law of motion;

The vertical component of the force;

$F_n = mgcos(\theta)$

The frictional force on the block;

$F_k = \mu_k F_n\\\\F_k = \mu_k mg\ cos(\theta)$

The horizontal component of the force;

$-mgsin(\theta) - F_k = ma\\\\-mgsin(\theta) - \mu_kmgcos(\theta) = ma\\\\-gsin(\theta) - \mu_kgcos(\theta) = a\\\\-g[sin(\theta) + \mu_kcos(\theta) ]= a\\\\-9.8[sin(31.5) + 0.368\times cos(31.5)] = a\\\\-9.8( 0.5225+ 0.3138) = a\\\\-8.20 \ m/s^2 = a$

The distance traveled by the block before stopping is calculated as;

$v_f^2 = v_0^2 + 2as$

when the block stops, the final velocity = 0

$0 = 15^2 + 2(-8.2)(s)\\\\0 = 225 - 16.4s\\\\16.4s = 225\\\\s = \frac{225}{16.4} \\\\s = 13.72 \ m$

Thus, the distance traveled by the block before stopping is 13.72 m.

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